(1) 證明:由題意得,
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=
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-a(x
1+x
2)-aln(x
1x
2),
f(
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)=
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-a(x
1+x
2)-2aln
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=
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-a(x
1+x
2)-aln
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∵
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-
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=
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>0(x
1≠x
2),∴
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>
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①
又∵0<x
1x
2<
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∴l(xiāng)nx
1x
2<ln
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∵a>0∴-alnx
1x
2>-aln
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②
由①②知
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>f(
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).
(2)(i)解:h(x)=
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=
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x
2-ax-alnx+
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ln
2x+a
2+
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.
∴h′(x)=x-a-
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+
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令F(x)=h′(x)=x-a-
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+
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,則y=F(x)在[1,+∞)上單調(diào)遞增.
∴F′(x)=
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,則當x≥1時,x
2-lnx+a+1≥0恒成立.
即x≥1時,a≥-x
2+lnx-1恒成立.
令G(x)=-x
2+lnx-1,則當x≥1時,G′(x)=
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<0.
∴G(x)=-x
2+lnx-1在[1,+∞)上單調(diào)遞減,從而G(x)
max=G(1)=-2.
故a≥G(x)
max=-2.即a的取值范圍是[-2,+∞).
(ii)證明::h(x)=
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x
2-ax-alnx+
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ln
2x+a
2+
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=a
2-(x+lnx)a+
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(x
2+ln
2x)+
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.
令P(a)=a
2-(x+lnx)a+
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(x
2+ln
2x),則P(a)=(a-
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)
2+
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≥
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.
令Q(x)=x-lnx,則Q′(x)=1-
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=
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.
顯然Q(x)在(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增,
則Q(x)
min=Q(1)=1,則P(a)≥
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.
故h(x)≥
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+
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=
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.
分析:(1)首先分別求出
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與f(
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);然后通過作差法或基本不等式等知識比較兩代數(shù)式中部分的大小;最后得出兩代數(shù)式整體的大�。�
(2)(i)首先求出h(x)及其導函數(shù)h′(x);然后根據(jù)y=h′(x)在[1,+∞)上單調(diào)遞增,得y=h′(x)的導函數(shù)大于等于0恒成立,則利用分離參數(shù)的方法可得關于a的不等式a≥-x
2+lnx-1(x≥1)恒成立;再運用導數(shù)法求出-x
2+lnx-1的最大值,此時a≥[-x
2+lnx-1]
max即可.
(ii)首先把h(x)表示成a為主元的函數(shù)h(x)=a
2-(x+lnx)a+
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(x
2+ln
2x)+
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;然后利用配方法得P(a)=a
2-(x+lnx)a+
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(x
2+ln
2x)=(a-
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)
2+
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≥
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;再通過構造函數(shù)Q(x)=x-lnx,并由導數(shù)法求其最小值進而得P(a)的最小值;最后得h(x)的最小值,即問題得證.
點評:本題主要考查函數(shù)單調(diào)性與導數(shù)的關系及最值與導數(shù)的關系,同時考查了不等式知識、比較法等;特別是導數(shù)法的連續(xù)運用是本題的難點.